Let $n\in\mathbb{N}$ be a positive integer. We say that an $n\times n$-matrix $A$ with all entries in $\{0,1\}$ is $k$-regular for some $k\in \{0,\ldots,n\}$ if the sum of every row and the sum of every column of $A$ equals $k$. Let $M(n, k)$ be the number of $k$-regular $n\times n$-matrices with all entries in $\{0,1\}$.

It's easy to see that $M(n,1)=n!$. Moreover, a symmetry argument shows that $M(n,k) = M(n, n-k)$ for all $k\in \{0,\ldots,n\}$.

Question. Given $n>1$, is it true that for all $k\in\{0,\ldots, n\}$ we have $M(n,k)\leq M(n, \lfloor\frac{n}{2}\rfloor)$?

$\begingroup$Worth pointing out: relevant articles are: A.Barvinok, On the number of matrices and a random matrix with prescribed row and column sums and 0-1 entries. Adv. Math. 224, No. 1, 316-339 (2010) and A. Barvinok, Matrices with prescribed row and column sums. Linear Algebra Appl. 436, No. 4, 820-844 (2012). Sadly, it seems (and I have looked at them very closely), while very interesting, the estimates proved in there are not precise enough to answer the question.$\endgroup$
– Peter HeinigMar 16 '18 at 17:12

2 Answers
2

I'm pretty sure this is unknown, though it would be great if I'm wrong.

Asymptotically there is a function $A(n)$ such that $M(n,n/2+t)\sim e^{-2t^2}A(n)$ for $t=o(n^{1/2})$, which follows from this paper of Canfield and McKay.

It is also known to be true for $n\le 20$.

To spell out the asymptotics a bit more, the paper cited above shows that
$$ M(n,k) = (e^{-1/2}+o(1)) \binom {n}{k}^{\!2n} \bigl( \lambda^\lambda(1-\lambda)^{1-\lambda}\bigr)^{n^2}, \qquad(*)$$
as $n\to\infty$, where $\lambda=k/n$ and $cn/\log n\le k\le n-cn/\log n$ for a particular constant $c$. The value $c=1/3$ will do. This expression without the error term is unimodal for $0\le k\le n$ and even with the error term it is unimodal if $n$ is large enough. The same formula holds for $k=o(n^{1/2})$ (McKay and Wang, 2003). A new method of Liebenau and Wormald will (I'm 100% confident) show that $(*)$ is also true for the intermediate ranges of $k$, but it is not published yet. Then we will know that $M(n,k)$ is unimodal in $k$ provided $n$ is large enough.

If we just want to know whether $M(n,k)\le M(n,\lfloor n/2\rfloor)$, as asked, and don't care about unimodality, then what remains asymptotically is to show that
$M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le \frac13 n/\log n$. This follows from the pairing (configuration) model for random $k$-regular bipartite graphs; namely
$$ M(n,k) \le \frac{ (nk)! }{ (k!)^{2n} }.\qquad(\#)$$
For large $k$, $(\#)$ is a pretty terrible bound, but if I didn't miscalculate it is sufficient to prove that $M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le \frac13 n/\log n$.

That completes the proof that $M(n,\lfloor n/2\rfloor)$ is the largest value if $n$ is large enough.

$\begingroup$What if we count ways of summing disjoint permutations? Can we estimate the numbers that way, at least towards determining unimodality? Gerhard "The Sum Of All Gears" Paseman, 2018.03.16.$\endgroup$
– Gerhard PasemanMar 16 '18 at 14:22

$\begingroup$@GerhardPaseman : The unimodality would follow if a precise estimate could be found for the expected number of permutation matrices covered by a random matrix in this class. The accuracy would need to be within a small constant. So far it has only been done asymptotically as far as I am aware.$\endgroup$
– Brendan McKayMar 16 '18 at 22:03

$\begingroup$Excellent progress, thanks for putting so much effort into this! I think I am going to accept this answer, because it answers the question for $n\in\mathbb{N}$ large enough$\endgroup$
– Dominic van der ZypenMar 17 '18 at 15:07

I can prove a related result. Perhaps someone can
modify the proof to solve Dominic's problem.

I use multivariate notation such as $x^\alpha=x_1^{\alpha_1}\cdots
x_m^{\alpha_m}$, where $\alpha=(\alpha_1,\dots,\alpha_m)$. Let
$\alpha\in \{0,1,2,\dots\}^m$ and $\beta\in\{0,1,2,\dots\}^n$. Let $f(\alpha,\beta)$ be the
number of $m\times n$ matrices with entries $0,1,2$,
and with each entry equal to 1 colored either red or blue, with row
sum vector $\alpha$ and column sum vector $\beta$.

Theorem.
$$ f(\alpha,\beta) \leq f((n,n,\dots,n),(m,m,\dots,m)). $$

Proof. Let $g(\alpha,\beta)$ be the
number of $m\times n$ matrices with entries $-2,0,2$,
and with each entry equal to 0 colored either red or blue, with row
sum vector $\alpha$ and column sum vector $\beta$. By dividing each
entry of such a matrix by 2 and then adding 1, it is clear that
$$ f(\alpha,\beta)=g\left(2\alpha-2(n,\dots,n),
2\beta-2(m,\dots,m)\right). $$
Hence we want to show that
$$ g(\alpha,\beta) \leq g((0,0,\dots,0),(0,0,\dots,0)). $$